There Is No Electric Field inside Tom - He's the Conductor!

Electric Field solutions:

p 496 - 498, Q 18, 19, P 33, 34, 35, 38, 43

1981B3 Solutions:
b) T cos(th) - mg = 0. T = mg/cos(th) = 5.65 x 10^-2 N

T sin(th) - qE = 0. E = T sin(th)/Q = 5650 N/C

c) Since the electric field will accelerate the sphere in the horizontal direction, and gravity will accelerate the ball in the vertical direction, the sphere will travel in a straight line directed downwards.

Questions:

18. The field decreases in strength A, B, and then C.

19. If field lines crossed, it would mean that a positive charge could move in two possible directions at the intersection of the lines. This contradicts the definition of the direction of electric field that we came up with, which was THE direction that a positive charge would move at a location.

33.

34.

35. Draw a FBD: Fnet = qE = ma. You know the mass of a proton from the back of the book, as well as the charge, so you can solve for E = 0.10 N/C.


38. The acceleration from the field comes from qE = ma, such that a = 3.24 x 10¹⁵ m/s^2.

We can use our constant acceleration equations because E, and therefore the electric force are constant. v^2 = vo^2 + 2a*delta_x.

v = 8.83 x 10⁶ m/s

b) The ratio of the forces qE/mg = 3.0 x 10^-15.

43. Again, from equilibrium, qE - mg = 0. E = (1.67 x 10^-27 kg)(9.8 m/s^2)/(1.6 x 10^-19C = 1.02 x 10^-7 N/C directed upwards.